Let A={x|x is  a  prime  number  and  x<30}.  The number of different rational numbers, whose numerator, and denominator belong to A, is

The set A contains 10 elements. Two different numbers for numerator and denominator from these can be obtained in 10×9=90  ways and each permutation will form a unique rational number different from one. In addition one will be formed if numerator and denominator are same. Hence, required number of numbers =90+1=91.